`int_0^1 (dx)/sqrt(1+x^2) =`____

Answer

To solve the integral

\[ \int_0^1 \frac{dx}{\sqrt{1+x^2}}, \]

we can follow these steps:

### Step 1: Identify the Integral
We need to evaluate the integral

\[ I = \int_0^1 \frac{dx}{\sqrt{1+x^2}}. \]

### Step 2: Use the Formula for Integration
We can use the formula for the integral

\[ \int \frac{dx}{\sqrt{a^2 + x^2}} = \ln |x + \sqrt{a^2 + x^2}| + C. \]

In our case, \(a^2 = 1\) (so \(a = 1\)), and we can rewrite our integral as:

\[ I = \int \frac{dx}{\sqrt{1^2 + x^2}}. \]

### Step 3: Apply the Limits of Integration
Using the formula, we get:

\[ I = \ln |x + \sqrt{1 + x^2}| \bigg|_0^1. \]

### Step 4: Evaluate at the Upper Limit
Now we evaluate at the upper limit \(x = 1\):

\[ I = \ln |1 + \sqrt{1 + 1^2}| = \ln |1 + \sqrt{2}|. \]

### Step 5: Evaluate at the Lower Limit
Next, we evaluate at the lower limit \(x = 0\):

\[ I = \ln |0 + \sqrt{1 + 0^2}| = \ln |0 + 1| = \ln 1 = 0. \]

### Step 6: Combine the Results
Now we combine the results from the upper and lower limits:

\[ I = \ln |1 + \sqrt{2}| - \ln |1| = \ln |1 + \sqrt{2}| - 0 = \ln(1 + \sqrt{2}). \]

### Final Answer
Thus, the value of the integral is:

\[ \int_0^1 \frac{dx}{\sqrt{1+x^2}} = \ln(1 + \sqrt{2}). \]

---